Formula for changing the momentum of a system. School encyclopedia
If on a body of mass m for a certain period of time Δ t force F → acts, then the body speed changes ∆ v → = v 2 → - v 1 → . We find that during the time Δ t the body continues to move with acceleration:
a → = ∆ v → ∆ t = v 2 → - v 1 → ∆ t .
Based on the fundamental law of dynamics, that is, Newton's second law, we have:
F → = m a → = m v 2 → - v 1 → ∆ t or F → ∆ t = m v 2 → - m v 1 → = m ∆ v → = ∆ m v → .
Definition 1Body impulse, or momentum is a physical quantity equal to the product of the mass of a body and the speed of its movement.
The momentum of a body is considered a vector quantity, which is measured in kilogram-meter per second (kg m/s).
Definition 2
Impulse force is a physical quantity equal to the product of a force and the time of its action.
Momentum is classified as a vector quantity. There is another formulation of the definition.
Definition 3
The change in the momentum of the body is equal to the impulse of the force.
When denoting momentum p → Newton's second law is written as:
F → ∆ t = ∆ p → .
This type allows us to formulate Newton's second law. Force F → is the resultant of all forces acting on the body. Equality is written as projections onto coordinate axes of the form:
F x Δ t = Δ p x ; F y Δ t = Δ p y ; F z Δ t = Δ p z .
Picture 1 . 16 . 1 . Body impulse model.
The change in the projection of the body's momentum onto any of the three mutually perpendicular axes is equal to the projection of the force impulse onto the same axis.
Definition 4
One-dimensional movement– this is the movement of a body along one of the coordinate axes.
Example 1
Using an example, consider the free fall of a body with an initial speed v 0 under the influence of gravity over a period of time t. When the O Y axis is directed vertically downward, the gravity impulse F t = mg, acting during time t, is equal to m g t. Such an impulse is equal to the change in the momentum of the body:
F t t = m g t = Δ p = m (v – v 0), whence v = v 0 + g t.
The recording coincides with the kinematic formula for determining the speed of uniformly accelerated motion. The magnitude of the force does not change over the entire interval t. When it is variable in magnitude, then the momentum formula requires substituting the average value of force F with p from the time interval t. Picture 1 . 16 . 2 shows how the impulse of a force that depends on time is determined.
Picture 1 . 16 . 2. Calculation of force impulse from the graph of dependence F (t)
It is necessary to select the interval Δ t on the time axis; it is clear that the force F(t) practically unchanged. Force impulse F (t) Δ t over a period of time Δ t will be equal to the area of the shaded figure. When dividing the time axis into intervals by Δ t i in the interval from 0 to t, add up the impulses of all acting forces from these intervals Δ t i , then the total impulse of force will be equal to the area of formation using the step and time axes.
By applying the limit (Δ t i → 0), you can find the area that will be limited by the graph F(t) and t axis. Using the definition of force impulse from a graph is applicable to any laws where there are changing forces and time. This decision leads to integration of function F(t) from the interval [ 0 ; t ] .
Picture 1 . 16 . 2 shows a force impulse located in the interval from t 1 = 0 s to t 2 = 10.
From the formula we find that F c p (t 2 - t 1) = 1 2 F m a x (t 2 - t 1) = 100 N s = 100 k g m / s.
That is, from the example we can see F with p = 1 2 F m a x = 10 N.
There are cases when determining the average force F c p is possible with known time and data on the reported impulse. With a strong impact on a ball with a mass of 0.415 kg g, a speed of v = 30 m/s can be reported. The approximate impact time is 8 10 – 3 s.
Then the momentum formula takes the form:
p = m v = 12.5 k g m/s.
To determine the average force F c p during an impact, it is necessary F c p = p ∆ t = 1.56 10 3 N.
We received very great importance, which is equal to a body weighing 160 kg.
When the movement occurs along a curved path, then the initial value p 1 → and the final
p 2 → can be different in magnitude and direction. To determine the momentum ∆ p →, a momentum diagram is used, where there are vectors p 1 → and p 2 →, and ∆ p → = p 2 → - p 1 → is constructed according to the parallelogram rule.
Example 2
As an example, see Figure 1. 16 . 2, where a diagram of the impulses of a ball bouncing off a wall is drawn. When serving, a ball with mass m with a speed v 1 → hits the surface at an angle α to the normal and rebounds with a speed v 2 → with an angle β. When hitting the wall, the ball was subjected to the action of a force F →, directed in the same way as the vector ∆ p →.
Picture 1 . 16 . 3. Rebounding of a ball from a rough wall and impulse diagram.
If a ball with mass m falls normally onto an elastic surface with a speed v 1 → = v → , then upon rebound it will change to v 2 → = - v → . This means that over a certain period of time the impulse will change and will be equal to ∆ p → = - 2 m v → . Using projections onto O X, the result will be written as Δ p x = – 2 m v x. From the drawing 1 . 16 . 3 it is clear that the O X axis is directed from the wall, then v x follows< 0 и Δ p x >0 . From the formula we find that the module Δ p is associated with the velocity module, which takes the form Δ p = 2 m v .
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1. As you know, the result of a force depends on its magnitude, point of application and direction. Indeed, the greater the force acting on the body, the greater the acceleration it acquires. The direction of the acceleration also depends on the direction of the force. So, by applying a small force to the handle, we can easily open the door, but if we apply the same force near the hinges on which the door hangs, then it may not be possible to open it.
Experiments and observations indicate that the result of a force (interaction) depends not only on the modulus of the force, but also on the time of its action. Let's do an experiment. We hang a load on a thread from the tripod, to which another thread is tied from below (Fig. 59). If you pull the lower thread sharply, it will break, and the load will remain hanging on the upper thread. If you now slowly pull the bottom thread, the top thread will break.
The impulse of force is a vector physical quantity equal to the product of force and the time of its action F t .
The SI unit of impulse of force is newton second (1 N s): [Ft] = 1 N s.
The force impulse vector coincides in direction with the force vector.
2. You also know that the result of a force depends on the mass of the body on which the force acts. So, than more mass body, the less acceleration it acquires under the action of the same force.
Let's look at an example. Let's imagine that there is a loaded platform on the rails. A carriage moving at some speed collides with it. As a result of the collision, the platform will acquire acceleration and move a certain distance. If a carriage moving at the same speed collides with a light trolley, then as a result of the interaction it will move a significantly greater distance than a loaded platform.
Another example. Let's assume that a bullet approaches the target at a speed of 2 m/s. The bullet will most likely bounce off the target, leaving only a small dent in it. If the bullet flies at a speed of 100 m/s, then it will pierce the target.
Thus, the result of the interaction of bodies depends on their mass and speed of movement.
The momentum of a body is a vector physical quantity equal to the product of the mass of the body and its speed.
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p = m v. |
The SI unit of momentum of a body is kilogram-meter per second(1 kg m/s): [ p] = [m][v] = 1 kg 1m/s = 1 kg m/s.
The direction of the body's momentum coincides with the direction of its speed.
Momentum is a relative quantity; its value depends on the choice of reference system. This is understandable, since speed is a relative quantity.
3. Let us find out how the impulse of force and the impulse of the body are related.
According to Newton's second law:
F = ma.
Substituting the expression for acceleration into this formula a= , we get:
F= , or
Ft = mv – mv 0 .
On the left side of the equation is the impulse of force; on the right side of the equality - the difference between the final and initial body impulses, t. e. change in the momentum of the body.
Thus,
the impulse of force is equal to the change in the momentum of the body.
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F t = D( m v). |
This is a different formulation of Newton's second law. This is exactly how Newton formulated it.
4. Let's assume that two balls moving on a table collide. Any interacting bodies, in this case balls, form system. Forces act between the bodies of the system: action force F 1 and counter force F 2. At the same time, the force of action F 1 according to Newton's third law is equal to the reaction force F 2 and is directed opposite to it: F 1 = –F 2 .
The forces with which the bodies of the system interact with each other are called internal forces.
In addition to internal forces, external forces act on the bodies of the system. Thus, the interacting balls are attracted to the Earth and are acted upon by the support reaction force. These forces are in this case external forces. During movement, the balls are subject to air resistance and friction. They are also external forces in relation to the system, which in this case consists of two balls.
External forces are forces that act on the bodies of a system from other bodies.
We will consider a system of bodies that is not affected by external forces.
A closed system is a system of bodies that interact with each other and do not interact with other bodies.
In a closed system, only internal forces act.
5. Let us consider the interaction of two bodies that make up a closed system. Mass of the first body m 1, its speed before interaction v 01, after interaction v 1 . Mass of the second body m 2, its speed before interaction v 02 , after interaction v 2 .
The forces with which bodies interact, according to the third law: F 1 = –F 2. The time of action of the forces is the same, therefore
F 1 t = –F 2 t.
For each body we write Newton’s second law:
F 1 t = m 1 v 1 – m 1 v 01 , F 2 t = m 2 v 2 – m 2 v 02 .
Since the left sides of the equalities are equal, then their right sides are equal, i.e.
m 1 v 1 – m 1 v 01 = –(m 2 v 2 – m 2 v 02).
Transforming this equality, we get:
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m 1 v 01 + m 1 v 02 = m 2 v 1 + m 2 v 2 . |
On the left side of the equation is the sum of the momenta of the bodies before the interaction, on the right is the sum of the momenta of the bodies after the interaction. As can be seen from this equality, the momentum of each body changed during interaction, but the sum of the impulses remained unchanged.
The geometric sum of the momenta of the bodies that make up a closed system remains constant for any interactions of the bodies of this system.
This is law of conservation of momentum.
6. A closed system of bodies is a model of a real system. There are no systems in nature that are not affected by external forces. However, in a number of cases, systems of interacting bodies can be considered closed. This is possible in the following cases: internal forces are much greater than external forces, interaction time is short, external forces compensate each other. In addition, the projection of external forces to any direction may be equal to zero, and then the law of conservation of momentum is satisfied for the projections of the impulses of interacting bodies to this direction.
7. Example of problem solution
Two railway platforms are moving towards each other at speeds of 0.3 and 0.2 m/s. The masses of the platforms are respectively 16 and 48 tons. At what speed and in what direction will the platforms move after automatic coupling?
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Given: |
SI |
Solution |
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v 01 = 0.3 m/s v 02 = 0.2 m/s m 1 = 16 t m 2 = 48 t v 1 = v 2 = v |
v 02 = v 02 = 1.6104kg 4.8104kg |
Let us depict in the figure the direction of movement of the platforms before and after interaction (Fig. 60). The gravity forces acting on the platforms and the support reaction forces cancel each other out. A system of two platforms can be considered closed |
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vx? |
and apply the law of conservation of momentum to it.
m 1 v 01 + m 2 v 02 = (m 1 + m 2)v.
In projections onto the axis X can be written:
m 1 v 01x + m 2 v 02x = (m 1 + m 2)v x.
Because v 01x = v 01 ; v 02x = –v 02 ; v x = – v, That m 1 v 01 – m 2 v 02 = –(m 1 + m 2)v.
Where v = – .
v= – = 0.75 m/s.
After coupling, the platforms will move in the direction in which the platform with the larger mass moved before the interaction.
Answer: v= 0.75 m/s; directed in the direction of movement of the cart with the greater mass.
Self-test questions
1. What is the impulse of a body?
2. What is called a force impulse?
3. How are the impulse of a force and the change in the momentum of a body related?
4. What system of bodies is called closed?
5. Formulate the law of conservation of momentum.
6. What are the limits of applicability of the law of conservation of momentum?
Task 17
1. What is the momentum of a body weighing 5 kg moving at a speed of 20 m/s?
2. Determine the change in momentum of a body weighing 3 kg in 5 s under the influence of a force of 20 N.
3. Determine the momentum of a car with a mass of 1.5 tons moving at a speed of 20 m/s in a reference frame associated with: a) a car stationary relative to the Earth; b) with a car moving in the same direction at the same speed; c) with a car moving at the same speed, but in the opposite direction.
4. A boy weighing 50 kg jumped from a stationary boat weighing 100 kg located in the water near the shore. At what speed did the boat move away from the shore if the boy’s speed is directed horizontally and is equal to 1 m/s?
5. A projectile weighing 5 kg, flying horizontally, exploded into two fragments. What is the speed of the projectile if a fragment weighing 2 kg at the explosion acquired a speed of 50 m/s, and a second fragment weighing 3 kg acquired a speed of 40 m/s? The velocities of the fragments are directed horizontally.
A 22-caliber bullet has a mass of only 2 g. If you throw such a bullet to someone, he can easily catch it even without gloves. If you try to catch such a bullet flying out of the muzzle at a speed of 300 m/s, then even gloves will not help.
If a toy cart is rolling towards you, you can stop it with your toe. If a truck is rolling towards you, you should move your feet out of its path.
Let's consider a problem that demonstrates the connection between a force impulse and a change in the momentum of a body.
Example. The mass of the ball is 400 g, the speed that the ball acquired after impact is 30 m/s. The force with which the foot acted on the ball was 1500 N, and the impact time was 8 ms. Find the impulse of force and the change in momentum of the body for the ball.
Change in body momentum
Example. Estimate the average force from the floor acting on the ball during impact.
1) During a strike, two forces act on the ball: ground reaction force, gravity.
The reaction force changes during the impact time, so it is possible to find the average reaction force of the floor.
2) Change in momentum 
body shown in the picture
3) From Newton's second law
The main thing to remember
1) Formulas for body impulse, force impulse;
2) Direction of the impulse vector;
3) Find the change in the momentum of the body
Derivation of Newton's second law in general form
Graph F(t). Variable force
The force impulse is numerically equal to the area of the figure under the graph F(t).
If the force is not constant over time, for example it increases linearly F=kt, then the momentum of this force is equal to the area of the triangle. You can replace this force with a constant force that will change the momentum of the body by the same amount in the same period of time
Average resultant force
LAW OF CONSERVATION OF MOMENTUM
Testing online
Closed system of bodies
This is a system of bodies that interact only with each other. There are no external forces of interaction.
IN real world There cannot be such a system; there is no way to remove all external interaction. A closed system of bodies is a physical model, just as a material point is a model. This is a model of a system of bodies that supposedly interact only with each other; external forces are not taken into account, they are neglected.
Law of conservation of momentum
In a closed system of bodies vector the sum of the momenta of the bodies does not change when the bodies interact. If the momentum of one body has increased, this means that at that moment the momentum of some other body (or several bodies) has decreased by exactly the same amount.
Let's consider this example. A girl and a boy are skating. A closed system of bodies - a girl and a boy (we neglect friction and other external forces). The girl stands still, her momentum is zero, since the speed is zero (see the formula for the momentum of a body). After a boy moving at a certain speed collides with a girl, she will also begin to move. Now her body has momentum. The numerical value of the girl’s momentum is exactly the same as how much the boy’s momentum decreased after the collision.
One body with a mass of 20 kg moves with a speed, a second body with a mass of 4 kg moves in the same direction with a speed of . What are the impulses of each body? What is the momentum of the system?
Impulse of a system of bodies is the vector sum of the momenta of all bodies included in the system. In our example, this is the sum of two vectors (since two bodies are considered) that are directed in the same direction, therefore
Now let's calculate the momentum of the system of bodies from the previous example if the second body moves in the opposite direction.
Since the bodies move in opposite directions, we obtain a vector sum of multidirectional impulses. Read more about vector sum.
The main thing to remember
1) What is a closed system of bodies;
2) The law of conservation of momentum and its application
Body impulse
The momentum of a body is a quantity equal to the product of the mass of the body and its speed.
It should be remembered that we're talking about about a body that can be represented as a material point. The momentum of the body ($p$) is also called the momentum. The concept of momentum was introduced into physics by René Descartes (1596–1650). The term “impulse” appeared later (impulsus in Latin means “push”). Momentum is a vector quantity (like speed) and is expressed by the formula:
$p↖(→)=mυ↖(→)$
The direction of the momentum vector always coincides with the direction of the velocity.
The SI unit of impulse is the impulse of a body with a mass of $1$ kg moving at a speed of $1$ m/s; therefore, the unit of impulse is $1$ kg $·$ m/s.
If a constant force acts on a body (material point) during a period of time $∆t$, then the acceleration will also be constant:
$a↖(→)=((υ_2)↖(→)-(υ_1)↖(→))/(∆t)$
where $(υ_1)↖(→)$ and $(υ_2)↖(→)$ are the initial and final velocities of the body. Substituting this value into the expression of Newton's second law, we get:
$(m((υ_2)↖(→)-(υ_1)↖(→)))/(∆t)=F↖(→)$
Opening the brackets and using the expression for the momentum of the body, we have:
$(p_2)↖(→)-(p_1)↖(→)=F↖(→)∆t$
Here $(p_2)↖(→)-(p_1)↖(→)=∆p↖(→)$ is the change in momentum over time $∆t$. Then the previous equation will take the form:
$∆p↖(→)=F↖(→)∆t$
The expression $∆p↖(→)=F↖(→)∆t$ is a mathematical representation of Newton's second law.
The product of a force and the duration of its action is called impulse of force. That's why the change in the momentum of a point is equal to the change in the momentum of the force acting on it.
The expression $∆p↖(→)=F↖(→)∆t$ is called equation of body motion. It should be noted that the same action - a change in the momentum of a point - can be achieved by a small force over a long period of time and great strength in a short period of time.
Impulse of the system tel. Law of Momentum Change
Impulse (amount of movement) mechanical system is called a vector equal to the sum of the momenta of all material points of this system:
$(p_(syst))↖(→)=(p_1)↖(→)+(p_2)↖(→)+...$
The laws of change and conservation of momentum are a consequence of Newton's second and third laws.
Let us consider a system consisting of two bodies. The forces ($F_(12)$ and $F_(21)$ in the figure with which the bodies of the system interact with each other are called internal.
Let, in addition to internal forces, external forces $(F_1)↖(→)$ and $(F_2)↖(→)$ act on the system. For each body we can write the equation $∆p↖(→)=F↖(→)∆t$. Adding the left and right sides of these equations, we get:
$(∆p_1)↖(→)+(∆p_2)↖(→)=((F_(12))↖(→)+(F_(21))↖(→)+(F_1)↖(→)+ (F_2)↖(→))∆t$
According to Newton's third law, $(F_(12))↖(→)=-(F_(21))↖(→)$.
Hence,
$(∆p_1)↖(→)+(∆p_2)↖(→)=((F_1)↖(→)+(F_2)↖(→))∆t$
On the left side there is a geometric sum of changes in the impulses of all bodies of the system, equal to the change in the impulse of the system itself - $(∆p_(syst))↖(→)$. Taking this into account, the equality $(∆p_1)↖(→)+(∆p_2) ↖(→)=((F_1)↖(→)+(F_2)↖(→))∆t$ can be written:
$(∆p_(syst))↖(→)=F↖(→)∆t$
where $F↖(→)$ is the sum of all external forces acting on the body. The result obtained means that the momentum of the system can only be changed by external forces, and the change in the momentum of the system is directed in the same way as the total external force. This is the essence of the law of change in momentum of a mechanical system.
Internal forces cannot change the total momentum of the system. They only change the impulses of individual bodies of the system.
Law of conservation of momentum
The law of conservation of momentum follows from the equation $(∆p_(syst))↖(→)=F↖(→)∆t$. If no external forces act on the system, then the right side of the equation $(∆p_(syst))↖(→)=F↖(→)∆t$ becomes zero, which means the total momentum of the system remains unchanged:
$(∆p_(syst))↖(→)=m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=const$
A system on which no external forces act or the resultant of external forces is zero is called closed.
The law of conservation of momentum states:
The total momentum of a closed system of bodies remains constant for any interaction of the bodies of the system with each other.
The result obtained is valid for a system containing an arbitrary number of bodies. If the sum of external forces is not equal to zero, but the sum of their projections to some direction is equal to zero, then the projection of the system’s momentum to this direction does not change. So, for example, a system of bodies on the surface of the Earth cannot be considered closed due to the force of gravity acting on all bodies, however, the sum of the projections of impulses on the horizontal direction can remain unchanged (in the absence of friction), since in this direction the force of gravity does not works.
Jet propulsion
Let us consider examples that confirm the validity of the law of conservation of momentum.
Let's take a children's rubber ball, inflate it and release it. We will see that when the air begins to leave it in one direction, the ball itself will fly in the other. The movement of the ball is an example jet propulsion. It is explained by the law of conservation of momentum: the total momentum of the “ball plus air in it” system before the air flows out is zero; it must remain equal to zero during movement; therefore, the ball moves in the direction opposite to the direction of flow of the jet, and at such a speed that its momentum is equal in magnitude to the momentum of the air jet.
Jet motion call the movement of a body that occurs when some part of it is separated from it at any speed. Due to the law of conservation of momentum, the direction of movement of the body is opposite to the direction of movement of the separated part.
Rocket flights are based on the principle of jet propulsion. Modern space rocket is very complex aircraft. The mass of the rocket consists of the mass of the working fluid (i.e., hot gases formed as a result of fuel combustion and emitted in the form of a jet stream) and the final, or, as they say, “dry” mass of the rocket remaining after the working fluid is ejected from the rocket.
When a jet gas stream high speed is thrown out of the rocket, the rocket itself rushes in the opposite direction. According to the law of conservation of momentum, the momentum $m_(p)υ_p$ acquired by the rocket must be equal to the momentum $m_(gas)·υ_(gas)$ of the ejected gases:
$m_(p)υ_p=m_(gas)·υ_(gas)$
It follows that the speed of the rocket
$υ_p=((m_(gas))/(m_p))·υ_(gas)$
From this formula it is clear that the speed of the rocket is greater, the greater more speed emitted gases and the ratio of the mass of the working fluid (i.e., the mass of the fuel) to the final (“dry”) mass of the rocket.
The formula $υ_p=((m_(gas))/(m_p))·υ_(gas)$ is approximate. It does not take into account that as the fuel burns, the mass of the flying rocket becomes less and less. The exact formula for rocket speed was obtained in 1897 by K. E. Tsiolkovsky and bears his name.
Work of force
The term “work” was introduced into physics in 1826 by the French scientist J. Poncelet. If in everyday life only human labor is called work, then in physics and, in particular, in mechanics it is generally accepted that work is performed by force. The physical quantity of work is usually denoted by the letter $A$.
Work of force is a measure of the action of a force, depending on its magnitude and direction, as well as on the movement of the point of application of the force. For a constant force and linear displacement, the work is determined by the equality:
$A=F|∆r↖(→)|cosα$
where $F$ is the force acting on the body, $∆r↖(→)$ is the displacement, $α$ is the angle between the force and the displacement.
The work of force is equal to the product of the moduli of force and displacement and the cosine of the angle between them, i.e., the scalar product of the vectors $F↖(→)$ and $∆r↖(→)$.
Work is a scalar quantity. If $α 0$, and if $90°
When several forces act on a body, the total work (the sum of the work of all forces) is equal to the work of the resulting force.
The unit of work in SI is joule($1$ J). $1$ J is the work done by a force of $1$ N along a path of $1$ m in the direction of action of this force. This unit is named after the English scientist J. Joule (1818-1889): $1$ J = $1$ N $·$ m. Kilojoules and millijoules are also often used: $1$ kJ $= 1,000$ J, $1$ mJ $= $0.001 J.
Work of gravity
Let us consider a body sliding along an inclined plane with an angle of inclination $α$ and a height $H$.
Let us express $∆x$ in terms of $H$ and $α$:
$∆x=(H)/(sinα)$
Considering that the force of gravity $F_т=mg$ makes an angle ($90° - α$) with the direction of movement, using the formula $∆x=(H)/(sin)α$, we obtain an expression for the work of gravity $A_g$:
$A_g=mg cos(90°-α) (H)/(sinα)=mgH$
From this formula it is clear that the work done by gravity depends on the height and does not depend on the angle of inclination of the plane.
It follows that:
- the work of gravity does not depend on the shape of the trajectory along which the body moves, but only on the initial and final position of the body;
- when a body moves along a closed trajectory, the work done by gravity is zero, i.e., gravity is a conservative force (forces that have this property are called conservative).
Work of reaction forces, is equal to zero, since the reaction force ($N$) is directed perpendicular to the displacement $∆x$.
Work of friction force
The friction force is directed opposite to the displacement $∆x$ and makes an angle of $180°$ with it, therefore the work of the friction force is negative:
$A_(tr)=F_(tr)∆x·cos180°=-F_(tr)·∆x$
Since $F_(tr)=μN, N=mg cosα, ∆x=l=(H)/(sinα),$ then
$A_(tr)=μmgHctgα$
Work of elastic force
Let an external force $F↖(→)$ act on an unstretched spring of length $l_0$, stretching it by $∆l_0=x_0$. In position $x=x_0F_(control)=kx_0$. After the force $F↖(→)$ ceases to act at point $x_0$, the spring is compressed under the action of force $F_(control)$.
Let us determine the work of the elastic force when the coordinate of the right end of the spring changes from $x_0$ to $x$. Since the elastic force in this area changes linearly, Hooke’s law can use its average value in this area:
$F_(control av.)=(kx_0+kx)/(2)=(k)/(2)(x_0+x)$
Then the work (taking into account the fact that the directions $(F_(control av.))↖(→)$ and $(∆x)↖(→)$ coincide) is equal to:
$A_(control)=(k)/(2)(x_0+x)(x_0-x)=(kx_0^2)/(2)-(kx^2)/(2)$
It can be shown that the form of the last formula does not depend on the angle between $(F_(control av.))↖(→)$ and $(∆x)↖(→)$. The work of elastic forces depends only on the deformations of the spring in the initial and final states.
Thus, the elastic force, like the force of gravity, is a conservative force.
Power power
Power is a physical quantity measured by the ratio of work to the period of time during which it is produced.
In other words, power shows how much work is done per unit of time (in SI - per $1$ s).
Power is determined by the formula:
where $N$ is power, $A$ is work done during time $∆t$.
Substituting into the formula $N=(A)/(∆t)$ instead of the work $A$ its expression $A=F|(∆r)↖(→)|cosα$, we obtain:
$N=(F|(∆r)↖(→)|cosα)/(∆t)=Fυcosα$
Power is equal to the product of the magnitudes of the force and velocity vectors and the cosine of the angle between these vectors.
Power in the SI system is measured in watts (W). One watt ($1$ W) is the power at which $1$ J of work is done for $1$ s: $1$ W $= 1$ J/s.
This unit is named after the English inventor J. Watt (Watt), who built the first steam engine. J. Watt himself (1736-1819) used a different unit of power - horsepower(hp), which he introduced so that the performance of a steam engine and a horse could be compared: $1$ hp. $= 735.5$ W.
In technology, larger power units are often used - kilowatt and megawatt: $1$ kW $= 1000$ W, $1$ MW $= 1000000$ W.
Kinetic energy. Law of change of kinetic energy
If a body or several interacting bodies (a system of bodies) can do work, then they are said to have energy.
The word “energy” (from the Greek energia - action, activity) is often used in everyday life. For example, people who can do work quickly are called energetic, having great energy.
The energy possessed by a body due to motion is called kinetic energy.
As in the case of the definition of energy in general, we can say about kinetic energy that kinetic energy is the ability of a moving body to do work.
Let us find the kinetic energy of a body of mass $m$ moving with a speed $υ$. Since kinetic energy is energy due to motion, its zero state is the state in which the body is at rest. Having found the work necessary to impart a given speed to a body, we will find its kinetic energy.
To do this, let’s calculate the work in the area of displacement $∆r↖(→)$ when the directions of the force vectors $F↖(→)$ and displacement $∆r↖(→)$ coincide. In this case the work is equal
where $∆x=∆r$
For the motion of a point with acceleration $α=const$, the expression for displacement has the form:
$∆x=υ_1t+(at^2)/(2),$
where $υ_1$ is the initial speed.
Substituting into the equation $A=F·∆x$ the expression for $∆x$ from $∆x=υ_1t+(at^2)/(2)$ and using Newton’s second law $F=ma$, we obtain:
$A=ma(υ_1t+(at^2)/(2))=(mat)/(2)(2υ_1+at)$
Expressing the acceleration through the initial $υ_1$ and final $υ_2$ velocities $a=(υ_2-υ_1)/(t)$ and substituting in $A=ma(υ_1t+(at^2)/(2))=(mat)/ (2)(2υ_1+at)$ we have:
$A=(m(υ_2-υ_1))/(2)·(2υ_1+υ_2-υ_1)$
$A=(mυ_2^2)/(2)-(mυ_1^2)/(2)$
Having now equated initial speed to zero: $υ_1=0$, we obtain an expression for kinetic energy:
$E_K=(mυ)/(2)=(p^2)/(2m)$
Thus, a moving body has kinetic energy. This energy is equal to the work that must be done to increase the speed of the body from zero to the value $υ$.
From $E_K=(mυ)/(2)=(p^2)/(2m)$ it follows that the work done by a force to move a body from one position to another is equal to the change in kinetic energy:
$A=E_(K_2)-E_(K_1)=∆E_K$
The equality $A=E_(K_2)-E_(K_1)=∆E_K$ expresses theorem on the change in kinetic energy.
Change in body kinetic energy(material point) for a certain period of time is equal to the work done during this time by the force acting on the body.
Potential energy
Potential energy is the energy determined by the relative position of interacting bodies or parts of the same body.
Since energy is defined as the ability of a body to do work, potential energy is naturally defined as the work done by a force, depending only on relative position tel. This is the work of gravity $A=mgh_1-mgh_2=mgH$ and the work of elasticity:
$A=(kx_0^2)/(2)-(kx^2)/(2)$
Potential energy of the body interacting with the Earth, they call a quantity equal to the product of the mass $m$ of this body by the acceleration of free fall $g$ and the height $h$ of the body above the Earth's surface:
The potential energy of an elastically deformed body is a value equal to half the product of the elasticity (stiffness) coefficient $k$ of the body and the squared deformation $∆l$:
$E_p=(1)/(2)k∆l^2$
The work of conservative forces (gravity and elasticity), taking into account $E_p=mgh$ and $E_p=(1)/(2)k∆l^2$, is expressed as follows:
$A=E_(p_1)-E_(p_2)=-(E_(p_2)-E_(p_1))=-∆E_p$
This formula allows you to give general definition potential energy.
The potential energy of a system is a quantity that depends on the position of the bodies, the change in which during the transition of the system from the initial state to the final state is equal to the work of the internal conservative forces of the system, taken with the opposite sign.
The minus sign on the right side of the equation $A=E_(p_1)-E_(p_2)=-(E_(p_2)-E_(p_1))=-∆E_p$ means that when work is performed by internal forces (for example, a fall bodies on the ground under the influence of gravity in the “rock-Earth” system), the energy of the system decreases. Work and changes in potential energy in a system always have opposite signs.
Since work determines only a change in potential energy, then only a change in energy has a physical meaning in mechanics. Therefore, the choice of the zero energy level is arbitrary and determined solely by considerations of convenience, for example, the ease of writing the corresponding equations.
Law of change and conservation of mechanical energy
Total mechanical energy of the system the sum of its kinetic and potential energies is called:
It is determined by the position of bodies (potential energy) and their speed (kinetic energy).
According to the kinetic energy theorem,
$E_k-E_(k_1)=A_p+A_(pr),$
where $A_p$ is the work of potential forces, $A_(pr)$ is the work of non-potential forces.
In turn, the work of potential forces is equal to the difference in the potential energy of the body in the initial $E_(p_1)$ and final $E_p$ states. Taking this into account, we obtain an expression for law of change of mechanical energy:
$(E_k+E_p)-(E_(k_1)+E_(p_1))=A_(pr)$
where the left side of the equality is the change in total mechanical energy, and the right side is the work of non-potential forces.
So, law of change of mechanical energy reads:
The change in the mechanical energy of the system is equal to the work of all non-potential forces.
A mechanical system in which only potential forces act is called conservative.
In a conservative system $A_(pr) = 0$. this implies law of conservation of mechanical energy:
In a closed conservative system, the total mechanical energy is conserved (does not change with time):
$E_k+E_p=E_(k_1)+E_(p_1)$
The law of conservation of mechanical energy is derived from Newton's laws of mechanics, which are applicable to a system of material points (or macroparticles).
However, the law of conservation of mechanical energy is also valid for a system of microparticles, where Newton’s laws themselves no longer apply.
The law of conservation of mechanical energy is a consequence of the uniformity of time.
Uniformity of time is that, under the same initial conditions, the occurrence of physical processes does not depend on at what point in time these conditions are created.
The law of conservation of total mechanical energy means that when the kinetic energy in a conservative system changes, its potential energy must also change, so that their sum remains constant. This means the possibility of converting one type of energy into another.
In accordance with various forms the movements of matter are considered different kinds energy: mechanical, internal ( equal to the amount kinetic energy of the chaotic movement of molecules relative to the center of mass of the body and the potential energy of interaction of molecules with each other), electromagnetic, chemical (which consists of the kinetic energy of the movement of electrons and electrical energy their interactions with each other and with atomic nuclei), nuclear, etc. From the above it is clear that the division of energy into different types Quite conditional.
Natural phenomena are usually accompanied by the transformation of one type of energy into another. For example, friction of parts of various mechanisms leads to the conversion of mechanical energy into heat, i.e. internal energy. In heat engines, on the contrary, internal energy is converted into mechanical energy; in galvanic cells, chemical energy is converted into electrical energy, etc.
Currently, the concept of energy is one of the basic concepts of physics. This concept is inextricably linked with the idea of the transformation of one form of movement into another.
This is how the concept of energy is formulated in modern physics:
Energy is a general quantitative measure of movement and interaction of all types of matter. Energy does not appear from nothing and does not disappear, it can only move from one form to another. The concept of energy links together all natural phenomena.
Simple mechanisms. Mechanism efficiency
Simple mechanisms are devices that change the magnitude or direction of forces applied to a body.
They are used to move or lift large loads with little effort. These include the lever and its varieties - blocks (movable and fixed), gates, inclined plane and its varieties - wedge, screw, etc.
Lever arm. Leverage rule
The lever is solid, capable of rotating around a fixed support.
The rule of leverage says:
A lever is in equilibrium if the forces applied to it are inversely proportional to their arms:
$(F_2)/(F_1)=(l_1)/(l_2)$
From the formula $(F_2)/(F_1)=(l_1)/(l_2)$, applying the property of proportion to it (the product of the extreme terms of a proportion is equal to the product of its middle terms), we can obtain the following formula:
But $F_1l_1=M_1$ is the moment of force tending to turn the lever clockwise, and $F_2l_2=M_2$ is the moment of force trying to turn the lever counterclockwise. Thus, $M_1=M_2$, which is what needed to be proven.
The lever began to be used by people in ancient times. With its help, it was possible to lift heavy stone slabs during the construction of pyramids in Ancient Egypt. Without leverage this would not be possible. After all, for example, for the construction of the Cheops pyramid, which has a height of $147$ m, more than two million stone blocks were used, the smallest of which weighed $2.5$ tons!
Nowadays, levers are widely used both in production (for example, cranes) and in everyday life (scissors, wire cutters, scales).
Fixed block
The action of a fixed block is similar to the action of a lever with equal arms: $l_1=l_2=r$. The applied force $F_1$ is equal to the load $F_2$, and the equilibrium condition is:
Fixed block used when you need to change the direction of a force without changing its magnitude.
Movable block
The moving block acts similarly to a lever, the arms of which are: $l_2=(l_1)/(2)=r$. In this case, the equilibrium condition has the form:
where $F_1$ is the applied force, $F_2$ is the load. The use of a moving block gives a double gain in strength.
Pulley hoist (block system)
A conventional chain hoist consists of $n$ moving and $n$ fixed blocks. Using it gives a gain in strength of $2n$ times:
$F_1=(F_2)/(2n)$
Power chain hoist consists of n movable and one fixed block. The use of a power pulley gives a gain in strength of $2^n$ times:
$F_1=(F_2)/(2^n)$
Screw
The screw is inclined plane, wound around the axis.
The equilibrium condition for the forces acting on the propeller has the form:
$F_1=(F_2h)/(2πr)=F_2tgα, F_1=(F_2h)/(2πR)$
where $F_1$ is the external force applied to the propeller and acting at a distance $R$ from its axis; $F_2$ is the force acting in the direction of the propeller axis; $h$ — propeller pitch; $r$ is the average thread radius; $α$ is the angle of inclination of the thread. $R$ is the length of the lever (wrench) rotating the screw with a force of $F_1$.
Efficiency
Coefficient useful action(efficiency) - the ratio of useful work to all expended work.
The efficiency factor is often expressed as a percentage and denoted Greek letter$η$ (“this”):
$η=(A_p)/(A_3)·100%$
where $A_n$ — useful work, $A_3$ is all the work expended.
Useful work always constitutes only a part of the total work that a person expends using one or another mechanism.
Part of the work done is spent on overcoming frictional forces. Since $A_3 > A_n$, the efficiency is always less than $1$ (or $< 100%$).
Since each of the works in this equality can be expressed as a product of the corresponding force and the distance traveled, it can be rewritten as follows: $F_1s_1≈F_2s_2$.
It follows that, winning with the help of a mechanism in force, we lose the same number of times along the way, and vice versa. This law is called the golden rule of mechanics.
The golden rule of mechanics is an approximate law, since it does not take into account the work of overcoming friction and gravity of the parts of the devices used. Nevertheless, it can be very useful in analyzing the operation of any simple mechanism.
So, for example, thanks to this rule, we can immediately say that the worker shown in the figure, with a double gain in the force of lifting the load by $10$ cm, will have to lower the opposite end of the lever by $20$ cm.
Collision of bodies. Elastic and inelastic impacts
The laws of conservation of momentum and mechanical energy are used to solve the problem of the motion of bodies after a collision: from the known impulses and energies before the collision, the values of these quantities after the collision are determined. Let us consider the cases of elastic and inelastic impacts.
An impact is called absolutely inelastic, after which the bodies form a single body moving at a certain speed. The problem of the speed of the latter is solved using the law of conservation of momentum of a system of bodies with masses $m_1$ and $m_2$ (if we are talking about two bodies) before and after the impact:
$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=(m_1+m_2)υ↖(→)$
It is obvious that the kinetic energy of bodies during an inelastic impact is not conserved (for example, for $(υ_1)↖(→)=-(υ_2)↖(→)$ and $m_1=m_2$ it becomes equal to zero after the impact).
An impact in which not only the sum of impulses is conserved, but also the sum of the kinetic energies of the impacting bodies is called absolutely elastic.
For an absolutely elastic impact, the following equations are valid:
$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=m_1(υ"_1)↖(→)+m_2(υ"_2)↖(→);$
$(m_(1)υ_1^2)/(2)+(m_(2)υ_2^2)/(2)=(m_1(υ"_1)^2)/(2)+(m_2(υ"_2 )^2)/(2)$
where $m_1, m_2$ are the masses of the balls, $υ_1, υ_2$ are the velocities of the balls before the impact, $υ"_1, υ"_2$ are the velocities of the balls after the impact.
Impulse(amount of motion) of a body is a physical vector quantity that is quantitative characteristics forward movement tel. The impulse is designated R. The momentum of a body is equal to the product of the mass of the body and its speed, i.e. it is calculated by the formula:
The direction of the impulse vector coincides with the direction of the body's velocity vector (directed tangent to the trajectory). The impulse unit is kg∙m/s.
Total momentum of a system of bodies equals vector the sum of the impulses of all bodies in the system:
Change in momentum of one body is found by the formula (note that the difference between the final and initial impulses is vector):
Where: p n – body impulse at the initial moment of time, p k – to the final one. The main thing is not to confuse the last two concepts.
Absolutely elastic impact– an abstract model of impact, which does not take into account energy losses due to friction, deformation, etc. No other interactions other than direct contact are taken into account. With an absolutely elastic impact on a fixed surface, the speed of the object after the impact is equal in magnitude to the speed of the object before the impact, that is, the magnitude of the impulse does not change. Only its direction can change. In this case, the angle of incidence equal to angle reflections.
Absolutely inelastic impact- a blow, as a result of which the bodies connect and continue their further movement as a single body. For example, when a plasticine ball falls on any surface, it completely stops its movement; when two cars collide, the automatic coupler is activated and they also continue to move further together.
Law of conservation of momentum
When bodies interact, the impulse of one body can be partially or completely transferred to another body. If a system of bodies is not acted upon by external forces from other bodies, such a system is called closed.
In a closed system, the vector sum of the impulses of all bodies included in the system remains constant for any interactions of the bodies of this system with each other. This fundamental law of nature is called law of conservation of momentum (LCM). Its consequences are Newton's laws. Newton's second law in momentum form can be written as follows:
As follows from this formula, if there is no external force acting on a system of bodies, or the action of external forces is compensated (the resultant force is zero), then the change in momentum is zero, which means that the total momentum of the system is conserved:
Similarly, one can reason for the equality of the projection of force on the selected axis to zero. If external forces do not act only along one of the axes, then the projection of the momentum onto this axis is preserved, for example:
Similar records can be made for other coordinate axes. One way or another, you need to understand that the impulses themselves can change, but it is their sum that remains constant. The law of conservation of momentum in many cases makes it possible to find the velocities of interacting bodies even when the values of the acting forces are unknown.
Saving momentum projection
Situations are possible when the law of conservation of momentum is only partially satisfied, that is, only when projecting onto one axis. If a force acts on a body, then its momentum is not conserved. But you can always choose an axis so that the projection of force on this axis is equal to zero. Then the projection of the impulse onto this axis will be preserved. As a rule, this axis is chosen along the surface along which the body moves.
Multidimensional case of FSI. Vector method
In cases where bodies do not move along one straight line, then in general case, in order to apply the law of conservation of momentum, you need to describe it along all coordinate axes involved in the problem. But solving such a problem can be greatly simplified if you use the vector method. It is used if one of the bodies is at rest before or after the impact. Then the law of conservation of momentum is written in one of the following ways:
From the rules for adding vectors it follows that the three vectors in these formulas must form a triangle. For triangles, the cosine theorem applies.
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How to successfully prepare for the CT in physics and mathematics?
In order to successfully prepare for the CT in physics and mathematics, among other things, it is necessary to fulfill three most important conditions:
- Study all topics and complete all tests and assignments given in the educational materials on this site. To do this, you need nothing at all, namely: devote three to four hours every day to preparing for the CT in physics and mathematics, studying theory and solving problems. The fact is that CT is an exam where it is not enough just to know physics or mathematics, you also need to be able to solve it quickly and without failures a large number of tasks for different topics and of varying complexity. The latter can only be learned by solving thousands of problems.
- Learn all the formulas and laws in physics, and formulas and methods in mathematics. In fact, this is also very simple to do; there are only about 200 necessary formulas in physics, and even a little less in mathematics. In each of these subjects there are about a dozen standard methods for solving problems of a basic level of complexity, which can also be learned, and thus solved completely automatically and without difficulty at the right time most CT. After this, you will only have to think about the most difficult tasks.
- Attend all three stages of rehearsal testing in physics and mathematics. Each RT can be visited twice to decide on both options. Again, on the CT, in addition to the ability to quickly and efficiently solve problems, and knowledge of formulas and methods, you must also be able to properly plan time, distribute forces, and most importantly, correctly fill out the answer form, without confusing the numbers of answers and problems, or your own last name. Also, during RT, it is important to get used to the style of asking questions in problems, which may seem very unusual to an unprepared person at the DT.
Successful, diligent and responsible implementation of these three points will allow you to show an excellent result at the CT, the maximum of what you are capable of.
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